Solve the equation: logx+log(x-2)=log(x+4).?

log(x) + log(x - 2) = log(x + 4)

log(x(x - 2)) = log(x + 4)

x(x - 2) = x + 4

x^2 - 2x = x + 4

x^2 - 3x - 4 = 0

(x - 4)(x + 1) = 0

x = 4 or x = -1

But, x = -1 is not a solution, since log(-1) and log(-3) are not defined. Thus, the one and only solution is x = 4.

When we're dealing with the same base, log M + log N = log (M*N)

So:

log x+log(x-2)=log(x+4)

log [x(x-2)] = log (x+4)

Therefore,

x(x - 2) = x + 4

x^2 - 2x = x + 4

x^2 - 3x - 4 = 0

(x - 4)(x + 1) = 0

x1 = 4; x2 = -1

We toss out x2 = -1 because in the original equation we see that log x would be log of a negative number (same case for log (x-2) but as long as you see one case of a negative log in the equation then that's enough to toss out that answer).

Your answer then is x = 4

We can verify the answer by substituting x = 4 in the original equation, and both sides must be the same:

log 4 + log (4 - 2) = log (4 + 4)

log 4 + log 2 = log 8

log (4*2) = log 8

log 8 = log 8

(I used the same rule again). I hope this helps...

log(x) + log(x - 2) = log(x + 4)

log(x(x - 2)) = log(x + 4)

x^2 - 2x = x + 4

x^2 - 3x - 4 = 0

(x - 4)(x + 1) = 0

x = 4, -1

However, use x = 4 since x = -1 causes a negative argument inside the log function.

rearrange using log rules log (x(x-2)/(x+4)) = 1

(x^2-2x)/(x+4) = 10

x(x-2) = x + 4

x^2 - 3x - 4 = 0

(x -4) (x + 1)

x = -1 (It's not a solution, is undefined)

x = 4 (This is the only solution)

4 or -1.

reject -1, x > 0.